PHYS
1441
–
Section
002
Lecture
#20
Wednesday
,
April 10, 2013
Dr.
Jae
hoon
Yu
•
Equations
of Rotational
Kinematics
•
Relationship
Between Angular and Linear
Quantities
•
Rolling
Motion of a Rigid
Body
•
Torque
•
Moment of Inertia
Announcements
•
Second non

comp term exam
–
Date and time: 4:00pm, Wednesday, April 17 in class
–
Coverage: CH6.1 through what we finish Monday, April 15
–
This exam could replace the first term exam if better
•
Remember that the lab final exams are next week!!
•
Special colloquium for 15 point extra credit!!
–
Wednesday, April 24, University Hall RM116
–
Class will be substituted by this colloquium
–
Dr.
Ketevi
Assamagan
from Brookhaven National Laboratory on
Higgs Discovery in ATLAS
–
Please mark your calendars!
!
Wednesday, April 10,
2013
2
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
3
f
Rotational Kinematics
The first type of motion we have learned in linear kinematics was
under a constant acceleration. We will learn about the rotational
motion under constant angular
acceleration (
α
),
because these are
the simplest motions in both cases.
Just like the case in linear motion, one can obtain
Angular
velocity
under constant
angular acceleration:
Angular displacement under
constant angular acceleration:
f
One can also obtain
f
2
0
0
0
2
2
f
0
t
0
t
1
2
t
2
v
Linear kinematics
Linear kinematics
f
x
2
1
0
2
o
x v t at
Linear kinematics
2
f
v
2
2
o f i
v a x x
o
v at
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
4
Rotational Kinematics Problem
Solving Strategy
•
Visualize the problem by drawing a picture.
•
Write down the values that are given for any of the
five
kinematic variables
and convert them to SI units.
–
Remember that the unit of the angle must be
in radians
!!
•
Verify that the information contains values for at least
three
of the five kinematic variables. Select the
appropriate equation.
•
When the motion is divided into segments, remember
that
the final angular
velocity of one segment is the
initial
velocity
for the next.
•
Keep in mind that there may be two possible answers
to a kinematics problem.
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
5
Example for Rotational Kinematics
A wheel rotates with a constant angular acceleration of 3.50 rad/s
2
. If
the angular speed of the wheel is 2.00 rad/s at
t
i
=0, a) through what
angle does the wheel rotate in 2.00s?
f
i
Using the angular displacement formula in the previous slide, one gets
t
2.00
2.00
1
2
3.50
2.00
2
11.0
r
ad
11.0
2
r
e
v
.
1.75
r
e
v
.
1
2
t
2
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
6
Example for Rotational Kinematics
cnt
’
d
What is the angular speed at t=2.00s?
f
i
t
Using the angular speed and acceleration relationship
Find the angle through which the wheel rotates between t=2.00 s
and t=3.00 s.
2
t
s
rad
/
00
.
9
00
.
2
50
.
3
00
.
2
3
t
2
10.8
r
ad
.
72
.
1
.
2
8
.
10
rev
rev
i
f
t
1
2
t
2
Using the angular kinematic formula
At t=2.00s
At t=3.00s
Angular
displacement
2
.
0
0
2
.
0
0
1
2
3
.
5
0
2
.
0
0
1
1
.
0
r
a
d
2
.
0
0
3
.
0
0
1
2
3
.
5
0
3
.
0
0
2
2
1
.
8
r
a
d
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
7
The blade is whirling with an angular velocity of +375
rad/s when
the
“
puree
”
button is pushed in.
When the
“
blend
”
button is pushed,
the blade accelerates and
reaches a
greater angular velocity after the blade has
rotated through an
angular displacement of +44.0 rad.
The angular acceleration has a
constant value of +1740
rad/s
2
.
Find the final angular velocity of the blade.
Ex. Blending with a Blender
θ
α
ω
ω
o
t
2 2
2
o
Which kinematic eq?
o
2
2
3
7
5
r
a
d
s
2
2
1
7
4
0
r
a
d
s
2
4
4
.
0
r
a
d
5
4
2
r
a
d
s
Which sign?
542rad s
Why?
Because the blade is accelerating in counter

clockwise!
+44.0rad
+375rad/s
?
+1740rad/s
2
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
8
Relationship Between Angular and Linear Quantities
What do we know about a rigid object that rotates
about a fixed axis of rotation?
When a point rotates, it has both the linear and angular
components in its motion.
What is the linear component of the motion you see?
v
Every particle (or
masslet
) in
an object
moves in a circle centered
at the same axis of rotation with the same angular velocity.
Linear
velocity along the tangential direction.
How do we related this linear component of the motion
with angular component?
l
r
The arc

length is
So the tangential speed
v
is
What does this relationship tell you about
the tangential speed of the points in the
object and their angular speed?
Although every particle in the object has the same
angular speed, its tangential speed differs and is
proportional to its distance from the axis of rotation.
The farther away the particle is from the center of
rotation, the higher the tangential speed.
The
direction
of
follows the
right

hand
rule.
l
t
r
t
r
t
r
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
9
Is
the
lion faster than
the
horse?
A rotating carousel has one child sitting on
the
horse near the outer edge
and another child on
the
lion halfway out from the center. (a) Which child
has the greater linear speed? (b) Which child has the greater angular
speed?
(a)
Linear speed is the distance traveled
divided by the time interval. So the child
sitting at the outer edge travels more
distance within the given time than the child
sitting closer to the center. Thus, the horse
is faster than the lion.
(b) Angular speed is the angle traveled divided by the time interval. The
angle both the children travel in the given time interval is the same.
Thus, both the horse and the lion have the same angular speed.
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
10
How about the acceleration?
v
t
Two
How many different linear acceleration components do
you see in a circular motion and what are they?
Total linear acceleration is
Since the tangential speed
v
is
What does this
relationship tell you?
Although every particle in the object has the same angular
acceleration, its tangential acceleration differs proportional to its
distance from the axis of rotation.
Tangential,
a
t
, and the radial acceleration,
a
r
.
a
t
The magnitude of tangential
acceleration
a
t
is
The radial or centripetal acceleration
a
r
is
r
a
What does
this tell you?
The father away the particle is from the rotation axis, the more radial
acceleration it receives. In other words, it receives more centripetal force.
a
v
t
f
v
t
0
t
r
f
r
0
t
r
v
2
r
2
r
2
r
a
t
2
a
r
2
r
2
r
2
2
r
2
4
r
r
f
0
t
r
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
11
A helicopter blade has an angular speed
of 6.50 rev/s and an
angular
acceleration of 1.30 rev/s
2
.
For point 1
on the blade, find
the magnitude of (a)
the
tangential speed and (b) the
tangential acceleration.
Ex. A Helicopter Blade
T
v
T
a
rev 2 rad
6.50
s 1 rev
40.8 rad s
r
3.00 m 40.8rad s 122m s
2
rev 2 rad
1.30
s 1 rev
2
8.17 rad s
2
3.00 m 8.17rad s
2
24.5m s
r
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
12
Rolling Motion of a Rigid Body
What is a rolling motion?
To simplify the discussion,
let’
s
make a few assumptions
Let
’
s consider a cylinder rolling on a flat surface, without slipping.
A more generalized case of a motion where the
rotational axis moves together with an object
Under what condition does this
“
Pure Rolling
”
happen?
The total linear distance the CM of the cylinder moved is
Thus the linear
speed of the CM is
A rotational motion about a moving axis
1.
Limit our discussion on very symmetric
objects, such as cylinders, spheres, etc
2.
The object rolls on a flat surface
R
s
s=Rθ
s
v
CM
s
t
The condition for a
“
Pure Rolling motion
”
R
t
R
R
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
13
More Rolling Motion of a Rigid Body
As we learned in rotational motion, all points in a rigid body
moves at the same angular speed but at different linear speeds.
At any given time, the point that comes to P has 0 linear
speed while the point at P
’
has twice the speed of CM
The magnitude of the linear acceleration of the CM is
A rolling motion can be interpreted as the sum of Translation and Rotation
CM
a
Why??
P
P
’
CM
v
CM
2v
CM
CM is moving at the same speed at all times.
P
P
’
CM
v
CM
v
CM
v
CM
+
P
P
’
CM
v=R
ω
v=0
v=Rω
=
P
P
’
CM
2v
CM
v
CM
CM
v
t
R
t
R
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
14
Starting from rest, the car accelerates
for 20.0 s
with a constant linear acceleration of 0.800 m/s
2
.
The radius of the tires is 0.330 m.
What is the
angle through which each wheel has rotated?
Ex. An Accelerating Car
θ
α
ω
ω
o
t
2
2 2
1
2
2.42rad s 20.0 s
a
r
2
2
0.800m s
2.42rad s
0.330 m
o
t
2
1
2
t
484 rad

2.42 rad/s
2
0
rad/s
20.0
s
?
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
15
Torque
Torque is the tendency of a force to rotate an object about an axis.
Torque,
, is a vector quantity.
Magnitude of torque is defined as the product of the force
exerted on the object to rotate it and the moment arm.
F
l
1
The line
of Action
Consider an object pivoting about the point
P
by
the force
F
being exerted at a distance
r
from
P
.
P
r
Moment arm
The line that extends out of the tail of the force
vector is called the
line of action.
The perpendicular distance from the pivoting point
P
to the
line of action
is called
the moment arm.
When there are more than one force being exerted on certain
points of the object, one can sum up the torque generated by each
force
vectorially
. The convention for sign of the torque is
positive
if rotation is in counter

clockwise
and
negative if clockwise
.
l
2
F
2
1 2
1 1 2 2
Fl F l
sin
F r
Fl
Magnitude of the Force
Lever Arm
Unit?
N m
Wednesday, April 10,
2013
PHYS 1441

002, Spring 2013
Dr. Jaehoon Yu
16
The tendon exerts a force of magnitude
790 N
on
the point P
. Determine the torque (magnitude
and
direction) of this force about the ankle joint
which is
located 3.6x10

2
m away from point P.
Ex. The Achilles Tendon
cos55
2
720 N 3.6 10 m cos55
790 N
F
2
3.6 10 m
2
3.6 10 cos55
2 2
3.6 10 sin 90 55 2.1 10
m
First, let’s find the lever arm length
So the torque is
Since the rotation is in clock

wise
3.6x10

2
m
7
2
0
N
3
.
6
1
0
2
m
s
i
n
3
5
1
5
N
m
15
N
m
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